(algebra) Proving subspaces- functions

[natalie:)]

Homework Statement

Is U={f E F($$\left|a,b\right|$$) f(a)=f(b)} a subspace of F($$\left| a,b \right|$$) where F($$\left| a,b \right|$$) is the vector space of real valued functions defined on the interval [a,b]?

Homework Equations

I know in order for something to be a subspace there are three conditions:
- existence of the zero vector $$\Theta$$
- closed under addition
- closed under scalar multiplication

The Attempt at a Solution

I tried to determine some boundaries for the function (not 1 to 1, all values will be positive, a and b can be any real number). I'm not really sure how to approach it ... i thought maybe if f(a)=0 then f(b) will also =0. But i don't really know how to prove that or the other two conditions.

I would really appreciate someone explaining the steps of how to solve this. Thanks!

 

[Mark44]

Homework Statement

Is U={f E F($$\left|a,b\right|$$) f(a)=f(b)} a subspace of F($$\left| a,b \right|$$) where F($$\left| a,b \right|$$) is the vector space of real valued functions defined on the interval [a,b]?

Homework Equations

I know in order for something to be a subspace there are three conditions:
- existence of the zero vector $$\Theta$$
- closed under addition
- closed under scalar multiplication

The Attempt at a Solution

I tried to determine some boundaries for the function (not 1 to 1, all values will be positive, a and b can be any real number). I'm not really sure how to approach it ... i thought maybe if f(a)=0 then f(b) will also =0.

Don't put any restrictions such as these on your functions.

But i don't really know how to prove that or the other two conditions.

I would really appreciate someone explaining the steps of how to solve this. Thanks!

Clearly U contains the zero function.

Now, let f and g be any arbitrary functions in U, and c any scalar. What can you say about (f + g)(a) and (f + g)(b)?

What can you say about cf(a) and cf(b)?

 

[natalie:)]

Oh I understand we can just name two functions since U represents the set of functions...
well (f+g)(a) would = (f+g)(b) and would be equivalent to f(a)+g(a)=f(b)+g(b)? I'm not really sure if I'm right about that.

then c*f(a)=c*f(b)..

I understand the point of what's happening but not the logic behind what's actually happening, if that makes any sense...

 

[Mark44]

Oh I understand we can just name two functions since U represents the set of functions...
well (f+g)(a) would = (f+g)(b) and would be equivalent to f(a)+g(a)=f(b)+g(b)? I'm not really sure if I'm right about that.

(f+g)(a) = f(a) + g(a) = f(b) + g(b) = (f + g)(b)
Can you give reasons for each of the = signs above?
What can you conclude about the function f + g?

then c*f(a)=c*f(b)..

Or better, (cf)(a) = c*f(a) = c*f(b) = (cf)(b)
Can you give reasons for each = above? What can you conclude about the function cf?

I understand the point of what's happening but not the logic behind what's actually happening, if that makes any sense...

 

[natalie:)]

Ok well...
(f+g)(a)=f(a)+g(a) i have no idea i want to say distributive property?
f(a)+g(a)= f(b)+g(b) because of the limitations in the question right? if they both have to come to the same solution.. f(a)=f(b) then g(a)=g(b). For some reason all i can think about is extreme value theorem which has nothing to do with this.

The function f+g is... no idea.

For scalar multiplication it makes sense that if you multiply any scalar by the function its the same as multiplying it inside the function c*f(a)=(cf)(a), and again I'm not really sure but think it has something to do with the f(a)=f(b)...

Does the absolute value change anything in solving or does it just imply the functions x-values stay above the axis?

Thank you for your patience!

 

[Mark44]

Ok well...
(f+g)(a)=f(a)+g(a) i have no idea i want to say distributive property?

No, this is the sum of two functions is defined. The distributive property shows how multiplication distributes over a sum of numbers. There is no multiplication in (f + g)(a).

f(a)+g(a)= f(b)+g(b) because of the limitations in the question right?

Sort of. We're assuming that f and g are in U, so f(a) = f(b) and g(a) = g(b). Adding g(a) to f(a) is the same as adding g(b) to f(b).

if they both have to come to the same solution.. f(a)=f(b) then g(a)=g(b).

No, one doesn't follow from the other. They both follow from our assumption that f and g are in U.

For some reason all i can think about is extreme value theorem which has nothing to do with this.

The function f+g is... no idea.

What you should conclude is that when f and g are in U, then their sum, f + g, will also be in U. Another way to say this is that U is closed under addition.

For scalar multiplication it makes sense that if you multiply any scalar by the function its the same as multiplying it inside the function c*f(a)=(cf)(a), and again I'm not really sure but think it has something to do with the f(a)=f(b)...

There is nothing going on "inside the function." c*f(a) = (cf)(a) by the definition of scalar multiples of a function.

Given two functions f and g, you can create four new functions: f + g, f - g, f*g, and f/g. Besides these four using arithmetic operations, there is also the composition f o g, where (f o g)(x) = f(g(x)). And there's the other order, g o f.

Does the absolute value change anything in solving or does it just imply the functions x-values stay above the axis?

What absolute value?

 

[natalie:)]

Wow I was really not understanding but now really am!
Oh and i was confused at the notation i thought the square brackets were abs value bars but they're brackets

Thank you so much for your help!

 

[Mark44]

Wow I was really not understanding but now really am!

Cool!

Oh and i was confused at the notation i thought the square brackets were abs value bars but they're brackets

Thank you so much for your help!

You're welcome. I enjoy doing it.